I am trying to find a conservative approximation for the propagated estimation error of a investment portfolio's variance (comprising two assets), given we know the estimation error for the variance of the two underlying assets and the correlation coefficient between these assets.

I have attempted to work through this using standard results for the propagation of estimation error but I am unsure that the end result is correct. Can you confirm whether you think this is a reasonable conservative estimation.

I use the assumptions that the asset weights will always be between 0 and 1, and furthermore that the error of the variance of the two assets will also always be between 0 and 1.

[\begin{array}{l} \sigma _p^2 = w_1^2\sigma _1^2 + w_2^2\sigma _2^2 + 2{w_1}{w_2}{\sigma _{1,2}}\\ {\sigma _{1,2}} = {\sigma _1}{\sigma _2}{\rho _{1,2}}\\ {\varepsilon _{{\sigma _1}}} = \frac{{{\varepsilon _{\sigma _1^2}}}}{{2{\sigma _1}}}\\ {\varepsilon _{{\sigma _2}}} = \frac{{{\varepsilon _{\sigma _2^2}}}}{{2{\sigma _2}}}\\ {\varepsilon _{{\sigma _{1,2}}}} \approx {\sigma _{1,2}}\left( {\frac{{{\varepsilon _{{\sigma _1}}}}}{{{\sigma _1}}} + \frac{{{\varepsilon _{{\sigma _2}}}}}{{{\sigma _2}}} + \frac{{{\varepsilon _{{\rho _{1,2}}}}}}{{{\rho _{1,2}}}}} \right) = {\sigma _{1,2}}\left( {\frac{{{\varepsilon _{\sigma _1^2}}}}{{2\sigma _1^2}} + \frac{{{\varepsilon _{\sigma _2^2}}}}{{2\sigma _2^2}} + \frac{{{\varepsilon _{{\rho _{1,2}}}}}}{{{\rho _{1,2}}}}} \right)\\ {\sigma _{{\varepsilon _{\sigma _1^2}},{\varepsilon _{{\sigma _{1,2}}}}}} = \frac{{{\sigma _{1,2}}\varepsilon _{\sigma _1^2}^2}}{{2\sigma _1^2}}\\ {\varepsilon _{\sigma _p^2}} \approx w_1^2\varepsilon _{\sigma _1^2}^{} + w_2^2\varepsilon _{\sigma _2^2}^{} + 2{w_1}{w_2}{\varepsilon _{{\sigma _{1,2}}}} + 2w_1^3{w_2}\left( {\frac{{{\sigma _{1,2}}\varepsilon _{\sigma _1^2}^2}}{{2\sigma _1^2}}} \right) + 2w_2^3{w_1}\left( {\frac{{{\sigma _{1,2}}\varepsilon _{\sigma _2^2}^2}}{{2\sigma _2^2}}} \right)\\ 0 \le {w_i},{\varepsilon _{{\sigma _j}}} \le 1 \Rightarrow w_i^3 \le {w_i},\varepsilon _{{\sigma _j}}^2 \le {\varepsilon _{{\sigma _j}}}\\ {\varepsilon _{\sigma _p^2}} \le w_1^2\varepsilon _{\sigma _1^2}^{} + w_2^2\varepsilon _{\sigma _2^2}^{} + 2{w_1}{w_2}{\sigma _{1,2}}\left( {\frac{{{\varepsilon _{\sigma _1^2}}}}{{\sigma _1^2}} + \frac{{{\varepsilon _{\sigma _2^2}}}}{{\sigma _2^2}} + \frac{{{\varepsilon _{{\rho _{1,2}}}}}}{{{\rho _{1,2}}}}} \right) = {w^T}\Sigma w\\ w = \left( {\begin{array}{*{20}{c}} {{w_1}}\\ {{w_2}} \end{array}} \right),\Sigma = \left( {\begin{array}{*{20}{c}} {{\varepsilon _{\sigma _1^2}}}&{{\sigma _{1,2}}\left( {\frac{{{\varepsilon _{\sigma _1^2}}}}{{\sigma _1^2}} + \frac{{{\varepsilon _{\sigma _2^2}}}}{{\sigma _2^2}} + \frac{{{\varepsilon _{{\rho _{1,2}}}}}}{{{\rho _{1,2}}}}} \right)}\\ {{\sigma _{1,2}}\left( {\frac{{{\varepsilon _{\sigma _1^2}}}}{{\sigma _1^2}} + \frac{{{\varepsilon _{\sigma _2^2}}}}{{\sigma _2^2}} + \frac{{{\varepsilon _{{\rho _{1,2}}}}}}{{{\rho _{1,2}}}}} \right)}&{{\varepsilon _{\sigma _2^2}}} \end{array}} \right) \end{array}]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.